Solving a System of Linear Equations - Buffalo State Collegemath.buffalostate.edu/~it/projects/rogers.pdf · Solving a System of Linear Equations ... Analysis, Inquiry, and Design

Solving a System of Linear Equations

Math A/Grade 9

5 Day Lesson Plan

Tools:1) TI 83 Plus Calculator

2) Overhead Projector with TI 83 View Screen 3) TI CBR

By

Glenn Rogers

I2T2

Summer 2004

Overall Objectives for the UnitOverall Objectives for the Unit

1) Students will be able to use readily available technology to solve systems of linear

equations.

2) Understand the meaning of "system of linear equations" and be able to explain it

in "real life" terms.

3) Students will be able to solve systems of linear equations both manually and with

the assistance of the TI 83 Plus.

4) Students will be able to solve systems of linear equations with the TI 83 Plus

graphically, algebraically, and with the use of matrices.

NYS Standards CoveredNYS Standards Covered

Standard 1: Analysis, Inquiry, and Design.

1) Abstraction and symbolic representation are used to communicate mathematically.

2) Critical thinking skills are used in the solution of mathematical problems.

Standard 2: Information Systems.

1) Information technology is used to retrieve, process, and communicate information

and as a tool to enhance learning.

Standard 3: Mathematics.

1) Students use mathematical reasoning to analyze mathematical situations, make

conjectures, gather evidence, and construct an argument.

2) Students use mathematical operations and relationships among them to

understand mathematics.

3) Students use mathematical modeling to provide a means of presenting,

interpreting, communicating, and connecting mathematical information and

relationships.

4) Students use measurement to provide a link between the abstraction of

mathematics and the real world.

Standard 7: Interdisciplinary Problem Solving.

1) The knowledge and skills of mathematics, science, and technology are used

together to make informed decisions and solve problems.

NCTM Standards CoveredNCTM Standards Covered

Algebra Standard.

1) Represent and analyze mathematical situations and structures using algebraic

symbols

2) Understand patterns, relations, and functions.

Problem Solving Standard.

1) Build new mathematical knowledge through problem solving

2) Solve problems that arise in mathematics and in other contexts.

3) Apply and adapt a variety of appropriate strategies to solve problems.

Communication Standard.

1) Communicate their mathematical thinking coherently and clearly to peers,

teachers, and others.

2) Use the language of mathematics to express mathematical ideas precisely.

Connections Standard.

1) Understand how mathematical ideas interconnect and build on one another to

produce a coherent whole.

Representation Standard.

1) Select, apply, and translate among mathematical representations to solve

problems.

Materials and Equipment NeededMaterials and Equipment Needed

Day One.

1) TI CBR

2) Class set of TI-83 plus calculators.

3) Overhead projector with TI-83 view screen.

Day Two.

1) Class work sheets

2) Take home work sheets.

Day Three.

1) TI-83 plus calculators.


Day Four.



Day Five.



Overview of Five Day Unit PlanOverview of Five Day Unit Plan

Day One: Introduction to system of linear equations.

1) What is a "system of linear equations"? Demonstration using TI-CBR.

2) Class discussion on what it means to solve a system of linear equations.

Day Two: Solving systems of linear equations manually.

1) Demonstrate to students the method of elimination and the method of substitution

for solving systems of linear equations.

2) Students work through class work sheet in groups.

Day Three: Solving systems of linear equations graphically with TI-83.

1) Demonstrate how to solve systems of linear equations graphically on the TI-83

plus.

3) Students work in groups with TI-83 plus handouts.

Day Four: Matrices with TI-83 to Solve systems of linear equations.

1) Demonstrate how to use TI-83 plus calculator to solve systems of linear

equations via matrices.

2) Students form groups and solve work sheet problems using matrices.

Day Five: Applying Systems of Linear Equations to Real Life.

1) Demonstrate how students can use their knowledge to solve everyday problems.

Day One: Introduction to system of linear equations.

Objectives:

1) Students will understand the meaning of system of linear equations.

2) Students will be able to relate a system of linear equations to real life situations.

3) Students will begin to think about methods for solving system of linear equations.

Lesson Plan:

The teacher will begin the lesson with a brief review of a linear equation. It is

assumed for the purpose of this lesson plan that the students will be familiar with linear

equations of the form y = ax + b. It will again be emphasized that graphically, a straight

line represents a linear equation.

The teacher will next create a system of linear equations using the TI-CBR unit in

conjunction with the TI-83 plus. The following as a step by step guided outline for the

teacher.

Step 1:

Set up the TI-83 plus so that it can be viewed on the overhead projector using the

TI-83 view screen. Connect the TI-83 to the CBR unit. Ask for two students to

volunteer to be the "walkers" in the demonstration.

Step 2:

On the TI-83 open the APPS screen and highlight option 5: CBL/CBR, (A). Next

press enter and highlight option 3: Ranger, (B). Next press enter twice to get to the main

menu screen and highlight 1: SETUP/SAMPLE, (C). For this demonstration set real time

to no, time(s) to 4, and units to feet and hit start now to get to screen (D).

A) B)

C) D)

Step 3:

Have the first volunteer stand approximately 8 feet from the CBR unit press enter

on the TI-83 and have the student walk towards the CBR at a constant rate. The student

will produce a linear graph similar to (E).

E)

Step 4:

Next we produce the line of best fit using the linear regression capabilities of the

TI-83. Use the STAT button and highlight CALC, 4: LinReg (ax + b), (F). Enter

through to get the LinReg screen, (G).

F) G)

Step 5:

Instruct the students to round the a and b values to the nearest tenth to come up

with the linear equation y = -1.2 x + 6.7. Enter this equation in the y = screen and graph

to get screen (H).

H)

Step 6:

We now repeat steps 2 – 5 accept this time the second volunteer stands in front of

the CBR unit and walks away from the unit to produce graph (I), Linear Regression (J),

and graph (K), using y = 1.1x + .9 as the linear equation.

I) J)

K)

Step 7:

The teacher can now explain to the students that we have created a "system of

linear equations", namely y = -1.2 x + 6.7, and y = 1.1x + .9. We next plot the linear

equations together on one graph as demonstrated in (L) and (M).

L) M)

Step 8:

The teacher next demonstrates how this "system of linear equations" can be

solved by finding the point of intersection. On the TI-83 his 2nd CALC, 5: intersect (N).

and follow screens (O), (P), (Q), and (R) to get the point of intersection x = 2.5 and y =

3.7 when rounded to the nearest tenth. Hence student A and student B crossed paths at

2.5 seconds and 3.7 feet away from the CBR unit.

N) O)

P) Q)

R)

Step 9:

The teacher will spend the remainder of the class time discussing what makes up a

system of linear equations. The teacher can ask the following questions to the students.

1) Will the system always have a solution? No, if there is no point of

intersection then there is no solution.

2) What does it mean if there is no point of intersection? The two lines must be

parallel.

3) What happens if the two lines coincide (overlapping lines)? This means there

is an infinite amount of solutions.

Exit:

Outline the rest of the week for the students. Let the students know that they will

be taught several different methods for solving systems of linear equations, both

manually and with the assistance of the TI-83.

Homework:

Assign reading from text if appropriate.

Day Two: Solving systems of linear equations manually.

Objectives:

1) Students will learn how to solve a system of linear equations using the method of

substitution.

2) Students will learn how to solve a system of linear equations using the method of

elimination.

Lesson Plan:

The teacher will begin the lesson with a brief review of what we mean by system

of linear equations. A brief review of day ones lesson should be sufficient. The

following is a step by step breakdown of day two.

Step 1: The method of substitution.

On the chalkboard or over head draw the following system of linear equations, x

+ y = 8, and x – y = 2. Explain to the students that the solution to the system can be

found by solving the first equation for y and then substituting the expression for y into the

second equation as follows.

1) First we solve x + y = 8 for y. We subtract x from both sides of the equation to

obtain y = 8-x.

2) Now we can substitute the expression y = 8-x into the equation x-y = 2.

x-(8-x) = 2, we can now distribute the negative sign to obtain the equation x-

8+x =2, now we can combine like terms to obtain the equation 2x-8 = 2. Next we

add 8 to both sides ending up with 2x = 10. Finally we divide both sides of the

equation by 2 to get x = 5.

3) Now that we know the x value is 5, we can substitute it into the equation x + y = 8

to get (5) + y = 8. Next we subtract 5 from both sides to obtain y = 3.

4) We now have the solution to the system of linear equations x + y = 8, and x – y =

2. We can express the solution as an ordered pair (x, y) = (5, 3). Explain to the

students that the solution could also have been found by originally solving for the

x variable and then substituting the x value into the second equation. Both ways

would arrive at the same unique solution.

Step 2: The method of elimination.

On the chalkboard or over head draw the following system of linear equations, x

+ y = 8, and x – y = 2. Let the students know that we will now solve the same system of

linear equations using the elimination method as follows.

1) The first step in solving by elimination is adding appropriate multiples of the

given equations in order to eliminate one of the variables, x or y. The equation

that we are left with can then be solved for the remaining variable using ordinary

algebra techniques.

2) Since the system of linear equations contains a positive y and a negative y there is

no need to use a multiple, we can simply add the two equations together like so.

x + y = 8 x - y = 2

2x =10

3) Solving the equation 2x = 10 by dividing both sides by 2 we obtain the equation

x = 5.

4) We can now substitute the x value into the second equation x = y = 8 to get (5) + y

= 8. Subtract 5 from both sides to obtain y =3. Hence the answer (x, y) = (5, 3).

It is now beneficial to show the class an example of a system of linear equations

when the x or y values do not immediately negate each other when added. Take for

example 7x + 2y = 5, and 2x + 3y = 16. Let the students know that they can pick either

variable to be eliminated first. In this instant the easiest choice would be to eliminate the

y variables. Multiplying both sides of 7x + 2y = 5 by 3 and both sides of 2x + 3y = 16

by -2, we get 21x + 6y = 15 and -4x – 6y = -32. We now observe that 6y and -6y will

cancel each other. We add both equations to obtain 17x = -17. Therefore x = -1. We can

now take either of the original equations and plug in the x value and then solve for y. 2(-

1) + 3y = 16, 3y = 18, therefore y =6, hence the answer (x, y) = (-1, 6).

Exit:

Take a few minutes to answer student questions. Have the students form groups

of four and work on classroom handout.

Homework:

Either give students the supplied handout or assign appropriate questions from

current text being used.

In-Class Activity: Solving Systems of Linear equations.

Solve the following problems using method of substitution.

1) x + y = 5, x – y = -1

2) 6x + 2y = 14, 3x + 2y = 8

Solve the following problems using method of elimination.

3) x + y = 12, x – y = 4

4) 3x + y = 16, 2x + y = 11

Homework: Solving Systems of Linear equations.

Solve the following problems using method of substitution.

1) 2x + 3y = -1

3x – 2y = -8

2) x + 3y = 5

3x – y = -5

3) x + y = 9

x -2y = -6

Solve the following problems using method of elimination.

4) 3x - 5y = 4

4x + 3y = 15

5) 3x + 4y = -11

7x – 5y = 3

Day Three: Solving systems of linear equations graphically with TI-83.

Objectives:

1) Students will learn how to solve a system of linear equations graphically using the

TI-83.

2) Students will interact in groups to solve problems with the TI-83.

Lesson Plan:

The teacher will begin the lesson with a brief review of material learned during

day two. The teacher will then demonstrate with the TI-83 and the overhead projector

with TI-83 view screen how to solve a system of linear equations graphically as follows.

Step 1:

Set up the TI-83 plus so that it can be viewed on the overhead projector using the

TI-83 view screen. Tell the students we are about to solve the system of linear equations

y = -x + 8, and y = x + 2 graphically using the TI-83 calculators. We first input are

equations into the y = screen as in figure (A). We next graph the equations to show the

point of intersection as in figure (B).

A) B)

We next hit 2nd CALC option 5: intersect as in figure (C). We then are prompted to mark

are first curve as in figure (D). We tell the students they want to mark the first curve

close to the point of intersection.

C) D)

We now are prompted for the second curve as in figure (E), again we tell students to

mark the curve close to the point of intersection. Next we are prompted to take a guess at

the point of intersection as in figure (F).

E) F)

After entering the guess the TI-83 will give us the (x, y) values as in figure (G).

G)

We can now verify the answer is correct by substituting the (x, y) values into the original

equations. (5) = - (3) + 8, 5 =5 so it checks. (5) = (3) + 2, 5 = 5 and again it checks,

therefore the solution is correct.

Exit:



Homework:

Either give students the supplied handout or assign appropriate questions from

current text being used.

In-Class Activity: Solving Systems of Linear equations using TI-83.

Solve the following problems graphically using TI-83.

1) 3x + 4y = 26, x – 3y = 0

2) 3x – 4y =2, x = 2(7 – y)

3) (1/2)x + (1/3)y = 8, (3/2)x + (-4/3)y = -4

4) 2x = 3y, (2/3)x (-1/2)y = 2

Homework: Solving Systems of Linear equations using TI-83.

Solve the following problems graphically using TI-83.

1) 4x + 3y = 5

5x + 3y = 2

2) 5x + 15y = 23

35x + 105y = 161

3) x + y = 9

x -2y = -6

4) 2x - y = 6

x – (1/2) y = 5

5) 3x - 4y = 5

5x + 4y = 3

Day Four: Solving systems of linear equations via Matrices with TI-83.

Objectives:

3) Students will learn how to solve a system of linear equations via matrices using

the TI-83.

4) Students will interact in groups to solve problems with the TI-83.

Lesson Plan:

The teacher will begin the lesson with a brief overview of how to put a system of

linear equations into matrix form. When solving a system of linear equations it is the

coefficients and constants that we are concerned with. A system of equations can be

represented, manipulated, and solved based solely on these coefficients and constants. It

is a simple matter to then write the coefficients and constant terms in matrix form.

The teacher will now write the following system of linear equations on the

blackboard, x + y = 8, x – y = 2. The teacher will next demonstrate that the system can

be written in matrix form by writing the resulting matrix on the board. Using the TI-83

and the TI-83 view screen the teacher will demonstrate two ways to solve this system of

linear equations using matrices.

Step 1: Using the inverse key.

The teacher demonstrates how to access the matrices functions by pressing 2nd

then the x inverse key to bring up window (A). Scroll over to the edit option and enter

matrix 1: [A] to get to the edit screen (B). The teacher explains to the students since we

have two rows and two columns in the coefficients matrix we need a 2 x 2 matrix. After

defining matrix [A] as 2 by 2, we then enter are coefficient values as in screen (B).

A) B)

After we have defined are matrix [A] we can exit to the home screen again enter the

matrix options, highlight matrix 1: [A] and enter to show students that we now have

matrix [A] defined as in screen (C).

C)

Using the same steps as above we can next great the constants matrix as a 2 by 1 matrix

as in screens (D), (E), and (F).

D) E)

F)

The teacher now explains in order to solve the system of linear equations we

simply take the inverse of matrix [A] by inputting [A] onto the home screen and then

pressing the x inverse key, and multiply it times matrix [B] to get the solution to the

system as demonstrated in screen (G).

G)

Step 1: Using the rref key.

The teacher next demonstrates how to solve the system of linear equations using

the rref key. For this method we make one 2 by 3 matrix as shown in screen (A). Again

the teacher brings up matrix [A] on the home screen to show the students the new matrix

as in screen (B).

A) B)

The teacher then enters the matrix home screen and scrolls to the math option andchooses option B:rref as in screen (C). Pressing enter will take you to the home screenwith rref ready to go. The teacher next demonstrates that all the student now needs to dois enter matrix [A] into the rref([A]) and enter to get the solution x = 5 and y = 3 as inscreen (D).

C) D)

The teacher can then explain that these are acceptable ways to answer a question that asks

for "only" an algebraic solution.

Exit:



Homework:

No homework given.


Solve the following problems using TI-83 matrix inverse method.

1) x + y = 5, x – y = -1

2) 6x + 2y = 14, 3x + 2y = 8

Solve the following problems using TI-83 matrix rref command.

3) x + y = 12, x – y = 4

4) 3x + y = 16, 2x + y = 11

Day Five: Applying Systems of Linear Equations to Real Life.

Objectives:

1) Students will learn how to apply a system of linear equations to real life situations

via word problems.

2) Students will interact in groups to solve problems manually and with the TI-83.

Lesson Plan:

The teacher will begin the lesson with a brief review of the previous methods for

solving systems of linear equations. The entire 5 day lesson plan will now be brought to

its conclusion by teaching the students how to use what they have learned in solving

every day problems in a real life setting.

The teacher will now show the students how to turn a word problem into a system

of linear equations by drawing the following examples on the overhead and explaining,

step by step, how to break the problem down.

The teacher will instruct the students that we are about to solve an investment

problem using one of our methods for solving systems of linear equations. The student

has $12,000 to invest in two funds that pay 9% and 11% in simple interest. The students

earn an annual interest of $1,180. How do we find out how much of the $12,000 was

invested at each rate?

Step 1: The students must first recognize that there are two unknowns, the

amount of money invested at 9% and the amount invested at 11%. The two unknowns

will now become the variables x and y respectively.

Step 2: The teacher will now ask the students to make the sentences into two

linear equations as follows.

1) (The amount of money invested at 9%) + ( The amount of money invested at

11%) = $12,000

2) (The amount of money invested at 9%) * 9% + (The amount of money invested

at 11%) * 11% = total interest of $1,180.

3) The students can now substitute the variables x and y for the sentences to get the

following system of linear equations.

4) x + y = $12,000

0.09x + 0.11y = $1,180

The word problem has now been translated into a system of equations that the students

can now solve using one of the methods they have learned over the previous four days of

instruction. For review purposes the teacher will now solve the problem on the overhead

via the method of substitution as follows.

1) Solve for y in equation 1. x + y = $12,000, so y = $12,000 – x.

2) Substitute the value for y into equation 2. 0.09x + 0.11($12,000 – x) =

$1,180.

3) Solve for x in the new equation. 0.09x + 0.11($12,000 – x) = $1,180. 0.09 +

$1,320 – 0.11x = $1,180. -0.02x = -$140. x = $7,000

4) Finally we substitute the value for x in the y equation from step 1. x + y =

$12,000. $7,000 + y = $12,000. y = $5,000

5) So the answer is the amount of money invested at 9% was $7,000, and the

amount of money invested at 11% was $5,000.

The teacher can solve the system of equations generated by the word problem using the

TI-83 and TI-83 view screen by either the graphing method or the matrices method to

show the students that the answers are the same. Depending on time constraints the

teacher may chose to solve another word problem on the overhead before letting the

students work on the class assignment.

Exit:



Homework:

No homework given.


Solve the following word problem using method of substitution.

1) Five hundred tickets were sold for a certain music concert. The tickets for the

adults and children sold for $7.50 and $4.00, respectively, and the total receipts

for the performance were $3,312.50. How many of each kind of ticket were sold?

Solve the following word problem graphically using TI-83.

2) At a quick lunch counter, 6 pretzels and 2 cups of soda cost $5.50. Four pretzels

and 2 cups of soda cost $4.00. Find the cost of a pretzel and the cost of a cup of

soda.

Solve the following word problem via matrices using TI-83.

3) A baseball manager bought 4 bats and 9 balls for $76.50. On another day, she

bought 3 bats and 1 dozen balls at the same prices and paid $81.00. How much

did she pay for each bat and each ball?

Answer Key for Class Assignments and Homework.

Day Two:

In-Class activity:

1) (2, 3)2) (2, 1)3) (8, 4)4) (5, 1)

Homework:

1) (-2, 1)2) (-1, 2)3) (4, 5)4) (3, 1)5) (-1, -2)

Day Three:

In-Class activity:

1) (6, 2)2) (6, 4)3) (8, 12)4) (6, 4)

Homework:

1) (-3, 17/3)2) Same line, infinite solutions.3) (4, 5)4) Parallel lines, no solution.5) (1, -1/2)

Day Four:

In-Class activity:

1) (2, 3)2) (2, 1)3) (8, 4)4) (5, 1)

Day Five:

In-Class activity:

1) 375 adult, 125 children.2) $.75 pretzel, $.50 soda.3) $9.00 bat, $4.50 ball.

Resources Used In Project

Essential Algebra: A Calculator Approach, Demona/Leitzel. Addison-Wesley

publishing company, pages 274-280, copyright 1989.

Integrated Mathematics: Course 1, by Dressler and Keenan. Amsco School

Publications, INC. Pages 615-630, copyright 1989.

http://www.sosmath.com/soe/SE2002/SE2002.html. Systems of equations in two

variables.

Documents

Solving a System of Linear Equations - Buffalo State Collegemath.buffalostate.edu/~it/projects/rogers.pdf · Solving a System of Linear Equations ... Analysis, Inquiry, and Design